Real numbers/Completeness/Nested intervals/Roots/Section

In a family of nested intervals, the length of the intervals are a decreasing null sequence. However, we do not require a certain velocity of the convergence. An interval bisection is a special kind of nested intervals, where the next interval is either the lower or the upper half of the preceding interval.

Definition

A sequence of closed intervals

$I_{n}=[a_{n},b_{n}],\,n\in \mathbb {N} ,$

in ${}\mathbb {R}$ is called (a sequence of) nested intervals, if ${}I_{n+1}\subseteq I_{n}$ holds for all ${}n\in \mathbb {N}$, and if the sequence of the lengths of the intervals, i.e.

${\left(b_{n}-a_{n}\right)}_{n\in \mathbb {N} },$

converges

to ${}0$.

Theorem

Suppose that ${}I_{n}$, ${}n\in \mathbb {N}$, is a sequence of nested intervals in ${}\mathbb {R}$. Then the intersection

$\bigcap _{n\in \mathbb {N} }I_{n}$

contains exactly one point

${}x\in \mathbb {R}$. Nested intervals determine a unique real number.

Proof

See exercise.

$\Box$

Theorem

For every nonnegative real number ${}c\in \mathbb {R} _{\geq 0}$ and every ${}k\in \mathbb {N} _{+}$ there exists a unique nonnegative real number ${}x$ fulfilling

${}x^{k}=c\,.$

Proof

We define recursively nested intervals ${}[a_{n},b_{n}]$. We set

${}a_{0}=0\,$

and we take for ${}b_{0}$ an arbitrary real number with ${}b_{0}^{k}\geq c$. Suppose that the interval bounds are defined up to index ${}n$, the intervals fulfil the containment condition and that

${}a_{n}^{k}\leq c\leq b_{n}^{k}\,$

holds. We set

${}a_{n+1}:={\begin{cases}{\frac {a_{n}+b_{n}}{2}},{\text{ if }}{\left({\frac {a_{n}+b_{n}}{2}}\right)}^{k}\leq c\,,\\a_{n}{\text{ else}},\end{cases}}\,$

and

${}b_{n+1}:={\begin{cases}{\frac {a_{n}+b_{n}}{2}},{\text{ if }}{\left({\frac {a_{n}+b_{n}}{2}}\right)}^{k}>c\,,\\b_{n}{\text{ else}}.\end{cases}}\,$

Hence one bound remains and one bound is replaced by the arithmetic mean of the bounds of the previous interval. In particular, the stated properties hold for all intervals and we have a sequence of nested intervals. Let ${}x$ denote the real number defined by this nested intervals according to fact. Because of exercise, we have

${}x=\lim _{n\rightarrow \infty }a_{n}=\lim _{n\rightarrow \infty }b_{n}\,.$

Due to fact  (2), we get

${}x^{k}=\lim _{n\rightarrow \infty }a_{n}^{k}=\lim _{n\rightarrow \infty }b_{n}^{k}\,.$

Because of the construction of the interval bounds and due to fact, this is ${}\leq c$ but also ${}\geq c$, hence ${}x^{k}=c$.

$\Box$

This uniquely determined number is denoted by ${}{\sqrt[{k}]{c}}$ or by ${}c^{1/k}$.